PHYS 142 Lecture Notes - Lecture 8: Capacitor, Electric Field

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7 Apr 2015

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Phys 142: electromagnetism and optics lecture 8 : electric potential. We define the electric potential v (or, for brevity, just the potential) as: The unit of electric potential is the joule per coulomb, which is called the volt v: 1 volt = 1 v = 1 j/c. The electric potential, like the electric field, is a property of the source charges. The electric potential inside a parallel plate capacitor: The electric potential inside a parallel-plate capacitor is v = es ; where s is the distance form the negative electrode. The potential difference vc, or voltage between the two capacitor plates is: If we know a capacitor"s voltage v and the distance between the plates d, then the electric field strength within the capacitor is: This implies that the units of electric field are volts per meter, or v/m. Previously, we have been using electric field units of newtons per coulomb.

Related Questions

To understand the meaning of capacitance and ways of calculating capacitance.

When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the conductor to ground, and the conductor will have a nonzero potential V relative to ground. If more charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this conductor: C=q/V.

Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are often called electrodes), and the relevant voltage is the voltage between these two electrodes.

This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallel-plate capacitor, which consists of two plates each of area A separated by a small distanced with air or vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in mind that the voltage difference is the line integral of the electric field between the plates.

Assume that charge -q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric field E between the plates?
What is the voltage V between the plates of the capacitor?

beigemouse1

A parallel plate capacitor is charged to a potential difference of100 V and disconnected from the source. A slab of dielectric isthen inserted between the plates. Which of the followingchanges?
a. the potential difference
b. the capacitance
c. the charge on the plates
d. 1 and 2
e. all of the above
f. none of the above

2. Consider inserting a slab of dielectric material is insertedbetween the plates of a charged capacitor. The electric field inthe dielectric is
a. largest near the positively charged plate.
b. the same everywhere.
c. largest near the negatively charged plate.
d. None of the above.

3.A certain amount of work is required to bring two point charges qnear each other until they are a distance d apart. If the magnitudeof the point charges were doubled to 2q, would the work requiredalso double?
a. yes
b. no
C. It depends on the signs of the charges.
D. None of the above.

1. True or False Charging a parallelplatecapacitor requires no work provided the charges placed on thetwoplates are equal and opposite. Explainyourresponse.

2. Capacitance of a parallel plate capacitor increasedecreaseor stay the same if

a. the charge on theplatedoubles

b. the potentialdifferenceacross plates is tripled

c. the distance between the platesishalved

d. area of the plates isdoubled

e electric field isreducedby half

PHYS 142 Lecture Notes - Lecture 8: Capacitor, Electric Field

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